Question

Explain binding energy. If 0.3% be the mass defect in a nuclear fusion reaction then how much energy will be released in fusion reaction of mass of 1 kg ?

Hint:

The formula \(E = mc^2\) shows that even a very small amount of mass can be converted into a tremendous amount of energy because the conversion factor, \(c^2\), is extremely large (\(9 \times 10^{16}\)). This is the principle behind nuclear power and nuclear weapons.

Answer 1

Step 1: Explanation of Binding Energy:
Mass Defect: It is observed that the mass of a stable nucleus is always slightly less than the sum of the masses of its constituent protons and neutrons (nucleons) when they are free. This difference in mass is called the mass defect (\(\Delta m\)). \[ \Delta m = [Zm_p + (A-Z)m_n] - M_{nucleus} \] where \(Z\) is the atomic number, \(A\) is the mass number, \(m_p\) is the mass of a proton, \(m_n\) is the mass of a neutron, and \(M_{nucleus}\) is the actual mass of the nucleus.

Binding Energy (B.E.): According to Einstein's mass-energy equivalence principle (\(E=mc^2\)), this "lost" mass (\(\Delta m\)) is converted into an equivalent amount of energy. This energy holds the nucleons together inside the nucleus and is known as the binding energy of the nucleus. \[ B.E. = \Delta m . c^2 \] Therefore, binding energy can be defined as the energy required to break a nucleus apart into its individual protons and neutrons. A higher binding energy per nucleon indicates a more stable nucleus.

Step 2: Understanding the Numerical Problem:
In a nuclear reaction (like fusion), energy is released if the total mass of the products is less than the total mass of the reactants. This loss of mass, which is the mass defect of the reaction, is converted into the released energy. We are given the percentage of the initial mass that is converted into energy.
Step 3: Calculating the Mass Defect:
Mass of the substance undergoing fusion, \( m = 1 \) kg.
The mass defect is given as 0.3% of this mass.
We calculate the actual mass defect, \(\Delta m\): \[ \Delta m = 0.3% \text{ of } 1 \, \text{kg} = \frac{0.3}{100} \times 1 \, \text{kg} \] \[ \Delta m = 0.003 \, \text{kg} \] Step 4: Calculating the Energy Released:
Now, we use Einstein's mass-energy equivalence formula to find the energy (E) released: \[ E = \Delta m . c^2 \] where \( c \) is the speed of light in vacuum, \( c = 3 \times 10^8 \) m/s. Substituting the values: \[ E = (0.003 \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \] \[ E = 0.003 \times (9 \times 10^{16}) \, \text{J} \] \[ E = 0.027 \times 10^{16} \, \text{J} \] Writing this in standard scientific notation: \[ E = 2.7 \times 10^{14} \, \text{J} \] Step 5: Final Answer:
The energy released in the fusion reaction of 1 kg of mass is \( 2.7 \times 10^{14} \) Joules.